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For solving differential equations, especially the ones of the form

g(x)dx = h(y)dy

we solve the equation by 'integrating on both sides' to reveal the solution.

Understanding this for 'differentiating the equation on both sides' is relatively easy. We know that we can formulate an alternative equation in terms of differentials for the original equation involved and come out with a new 'differential' equation that holds because of the properties of the differentials.

But how does it work for Integration on both sides?? Am I missing any point here?? I have referred to multiple books but none give a satisfactory explanation. Integrating an equation on both sides seems really 'wrong', if I may dare to use the word.

Please help. I'm stuck with this thing and I can only begin to understand differential equations once this is cleared from my head.

Thank you very much !

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2 Answers 2

up vote 4 down vote accepted

The original equation was presumably $$h(y)\frac{dy}{dx}=g(x),$$ or something equivalent to this.

You are given the mysterious rule about "splitting" $\frac{dy}{dx}$. You probably were told at one time that $\frac{dy}{dx}$ is not a fraction, and now all of a sudden we are treating it as a fraction!

So let us not split it. Suppose that $H(y)$ is an antiderivative of $h(y)$, that is, a function whose derivative wiith respect to $y$ is $h(y)$. Let $G(x)$ be a function whose derivative with respect to $x$ is $g(x)$.

We recognize $h(y)\frac{dy}{dx}$ as the derivative with respect to $x$ of $H(y)$ (Chain Rule). So our equation can be written as $$\frac{d}{dx} H(y)=\frac{d}{dx}G(x).$$

Thus $H(y)$ and $G(x)$ have the same derivative with respect to $x$. So they differ by a constant, and we find $$H(y)=G(x)+C.$$

Now the important part: this is exactly what we get when we "split" $\frac{dy}{dx}$ and integrate on both sides. So whether or not the splitting and integrating makes sense, it gives the right answer.

If you wish, splitting and integrating can be treated as a senseless mnemonic that works, a "shortcut" to the real calculation using the Chain Rule. In fact, the individual terms $dy$ and $dx$ can be given meaning, but it is a little complicated. And Applied (and less Applied) people have an essentially correct intuition based on adding up "infinitely small" quantities. Unfortunately, it takes considerable effort to make that intuition rigorous.

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Hi. Can you point me to the source from where you learnt this explanation?? I refer Thomas' Calculus and Erwin Kreyszig for my engineering mathematics. Both these books donot have the explanation you have given... –  Raghavendra Kumar Aug 18 '12 at 6:43

The (separable) differential equation you are interested in, should be written in the form $$g(x) = h(y) y'(x).\qquad (1)$$

The first equation should be what your form (in terms of differentials) means and in which sense it is equivalent to (1) (I believe that this has been answered several times on this site, otherwise see differential of a function).

It is an easy exercise to check that the implicit equation $$\int_{x_0}^x\!dx\,g(x) = \int_{y_0}^{y(x)}\!dy\,h(y)\qquad (2)$$ solves (1) with the initial condition $y(x_0)= y_0$:

  • regarding the initial condition: setting $x=x_0$ we obtain $0 = \int_{y_0}^{y(x_0)}\!dy\,h(y)$ which is solved with $y(x_0)=y_0$ (but of course also potentially other solutions; nonlinear equations might have multiple solutions)

  • proof that (2) is a solution of (1): we take the derivative with respect to $x$ on both sides of (2). On the left side we obtain (with the Fundamental theorem of calculus) $g(x)$. On the left side we obtain (using the chain rule) $h(y) y'(x)$ which proves the equivalence of (2) and (1)

So in the end, saying that integrating both sides of your equation solves the differential equation (1) can be seen as a way to memories form of the solution (2).

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Hi Fabian, Can you elaborate a bit more on the reasoning behind the implicit equation? –  Raghavendra Kumar Aug 14 '12 at 5:48
I am not sure what you want to hear. The reasoning is that it solves the (separable) differential equation as explained in the post. –  Fabian Aug 14 '12 at 5:50
Dear Fabian, I did not understand the first bullet point you made after the eq(2). –  Raghavendra Kumar Aug 14 '12 at 5:56
The first bullet just shows that the solution (2) implements the initial condition $y(x_0)= y_0$. So I set $x=x_0$ and observe that $y=y_0$. –  Fabian Aug 14 '12 at 7:06

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